Problem: Let $a(x)=3x^3-4x^2+x$, and $b(x)=x^2+2$. When dividing $a$ by $b$, we can find the unique quotient polynomial $q$ and remainder polynomial $r$ that satisfy the following equation: $\dfrac{a(x)}{b(x)}=q(x) + \dfrac{r(x)}{b(x)}$, where the degree of $r(x)$ is less than the degree of $b(x)$. What is the quotient, $q(x)$ ? $ q(x)=$ What is the remainder, $r(x)$ ? $r(x)=$
Solution: Note that $a(x)$ has a higher degree than $b(x)$. This allows us to find a non-zero quotient polynomial, $q(x)$. [Why is this important?] Let's use long division with polynomials in order to find the quotient, $q(x)$ and remainder, $r(x)$ of $\ \dfrac{a(x)}{b(x)}=\dfrac{3x^3-4x^2+x}{x^2+2}$ : First, we divide ${x^2}$ into ${3x^3}$ and get ${3x}$ : $ \hphantom{1567|14} {3x}\\ {{{x^2}+2}}|\overline{{3x^3}-4x^2+x}\\ \hphantom{37.....|}\llap{-}\underline{(3x^3+0x^2+6x)}\\ \hphantom{37|3..........}-4x^2-5x\\ $ [What did we do here?] Next, we divide ${x^2}$ into ${-4x^2}$ to get ${-4}$ : $ \hphantom{1567|14} {3x \ {- \ \ 4}}\\ {{{x^2}+2}}|\overline{3x^3-4x^2+x}\\ \hphantom{37.....|}\llap{-}\underline{(3x^3+0x^2+6x)}\\ \hphantom{37|3...........}{-4x^2}-5x+0\\ \hphantom{37.............|}\llap{-}\underline{(-4x^2+0x-8)}\\ \hphantom{37|3.....................}{-5x+8}\\ $ [What did we do here?] The process stops here because $x^2+2$ is a polynomial of the second degree and $-5x+8$ is a polynomial of the first degree. So it follows that ${r(x)}={-5x+8}$, ${q(x)}={3x-4}$, and $ \dfrac{3x^3-4x^2+x}{x^2+2}={3x-4}+\dfrac{{-5x+8}}{x^2+2}$ To conclude, $q(x)=3x-4$ $r(x)=-5x+8$